Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 3 - Review Exercises - Page 300: 3

Answer

Standard form: $f(x)=(x-1)^2-1$ Vertex: $(1,-1)$ Axis of symmetry: $x=1$ The x-intercepts: $(2,0)$ and $(0,0)$
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Work Step by Step

$f(x)=x^2-2x~~$ ($a=1,~b=-2,~c=0$) $f(x)=x^2-2x+1-1$ $f(x)=x^2-2(1)x-1^2-1$ $f(x)=(x-1)^2-1$ $-\frac{b}{2a}=-\frac{-2}{2(1)}=1$ $f(1)=(1-1)^2-1=-1$ Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(1,-1)$ So, the axis of symmetry is $x=-\frac{b}{2a}=1$. The x-intercepts: $f(x)=0$ $(x-1)^2-1=0$ $(x-1)^2=1$ $x-1=±1$ $x=2$ or $x=0$ So, the x-intercept points are: $(2,0)$ and $(0,0)$
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